Question

Two circles P and Q, each of radius 2 cm, pass through each other’s centres. They intersect at points A and B. A circle R is drawn with diameter AB. What is the area of overlap (in square cm) between the circles R and P?

• A. \(\frac{8\pi}{3}-2 \sqrt{3} \)
• B. \(\frac{8\pi}{3} \)
• C. \(\frac{8\pi}{3}- \sqrt{3}\)
• D. \(\frac{17\pi}{6}- 2\sqrt{3}\)
• E. \(\frac{17\pi}{6}- \sqrt{3}\)

Answer 1

Answer: (E)

Step 1: Geometry of Circles P and Q 
- Circles P and Q each have radius \(2 \, \text{cm}\). - Since each passes through the other’s centre, the distance between P and Q is also \(2 \, \text{cm}\). - The line \(AB\) is the common chord of equal intersecting circles. - Length of chord \(AB\) can be calculated using Pythagoras in triangle formed by radius and half-chord.

Step 2: Length of AB
For two equal circles of radius \(r = 2\) cm, and distance between centres = \(r = 2\): \[ AB = 2 \sqrt{r^2 - \left(\frac{d}{2}\right)^2} \] where \(d = 2\). So, \[ AB = 2 \sqrt{2^2 - 1^2} = 2\sqrt{3} \]

Step 3: Radius of Circle R
Circle R is drawn on diameter \(AB\). So radius of R = \( \tfrac{AB}{2} = \sqrt{3}\).

Step 4: Overlap of Circle R and Circle P
Now, radius of P = 2, radius of R = \(\sqrt{3}\). Distance between their centres = \( \sqrt{2^2 - (\tfrac{AB}{2})^2} = 1 \). So the problem reduces to finding the intersection area of two circles of radii \(2\) and \(\sqrt{3}\), with centre distance 1.

Step 5: Formula for Overlap Area of Two Circles
If two circles of radii \(r_1, r_2\) intersect with distance \(d\) between centres, overlap area = \[ r_1^2 \cos^{-1}\!\left(\frac{d^2 + r_1^2 - r_2^2}{2 d r_1}\right) + r_2^2 \cos^{-1}\!\left(\frac{d^2 + r_2^2 - r_1^2}{2 d r_2}\right) - \tfrac{1}{2} \sqrt{(-d + r_1 + r_2)(d + r_1 - r_2)(d - r_1 + r_2)(d + r_1 + r_2)} \] Substitute \(r_1 = 2\), \(r_2 = \sqrt{3}\), \(d = 1\). The simplification leads to: \[ \text{Overlap area} = \tfrac{17\pi}{6} - \tfrac{3}{6} \]

Final Answer:
\[ \boxed{\tfrac{17\pi}{6} - \tfrac{3}{6}} \]